Katholieke Universiteit Leuven Department of Computer Science

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1 Extensions of Fibonacci lattice rules Ronald Cools Dirk Nuyens Report TW 545, August 2009 Katholieke Universiteit Leuven Department of Computer Science Celestijnenlaan 200A B-3001 Heverlee (Belgium

2 Extensions of Fibonacci lattice rules Ronald Cools Dirk Nuyens Report TW 545, August 2009 Department of Computer Science, KULeuven Abstract We study the trigonometric degree of pairs of embedded cubature rules for the approximation of two-dimensional integrals, where the basic cubature rule is a Fibonacci lattice rule The embedded cubature rule is constructed by simply doubling the points which results in adding a shifted version of the basic Fibonacci rule An explicit expression is derived for the trigonometric degree of this particular extension of the Fibonacci rule based on the index of the Fibonacci number Keywords : Fibonacci lattice rule, cubature, embedded rule MSC : Primary : 65D32

3 Extensions of Fibonacci Lattice Rules Ronald Cools and Dirk Nuyens Abstract We study the trigonometric degree of pairs of embedded cubature rules for the approximation of two-dimensional integrals, where the basic cubature rule is a Fibonacci lattice rule The embedded cubature rule is constructed by simply doubling the points which results in adding a shifted version of the basic Fibonacci rule An explicit expression is derived for the trigonometric degree of this particular extension of the Fibonacci rule based on the index of the Fibonacci number Dedicated to Ian Sloan s 70th birthday 1 Introduction We consider the approximation of integrals I[ f ] := f (xdx [0,1 s by weighted sums of function values In practice one wants more than one such approximation to obtain information on the accuracy of the approximation In order to approximate an integral together with an error estimate, one often uses two approximations Q 1 [ f ] and Q 2 [ f ] Then Q 1 [ f ] Q 2 [ f ] can be used as an approximation of the error of the less precise rule (In practice it is often used as an estimate of the Ronald Cools Department of Computer Science, KULeuven, Leuven, Belgium, ronaldcools@cskuleuvenbe, ronald Dirk Nuyens Department of Computer Science, KULeuven, Leuven, Belgium, and School of Mathematics & Statistics, University of NSW, Sydney, Australia, dirknuyens@cskuleuvenbe, dirkn 1

4 2 Ronald Cools and Dirk Nuyens error of the most expensive rule, while hoping that this is the most precise Practical robust error estimates are based on more than one such combinations Given a cubature formula Q 1 [ f ] = N j=1 w j f (x j for the approximation of an integral I[ f ], we are interested in a cubature formula Q 2 [ f ] = N j=1 w j f (x j + N+M j=n+1 w j f (x j that reuses the function evaluations of Q 1 and is better The quality criterion used in this paper is the trigonometric degree, and the cubature formulas will be lattice rules For other criteria, see [4] A cubature formula of trigonometric degree d integrates correctly trigonometric polynomials of degree d Specifically in s dimensions, it integrates exp(2πi r x correctly for all r = (r 1,r 2,,r s Z s that satisfy r := s k=1 r k d Embedded pairs of quadrature and cubature formulas of algebraic degree were already studied a long time ago (see, eg, [6, 2, 3] We are unaware of an attempt to do this for the trigonometric case This paper describes a first attempt, limited to the 2-dimensional case More examples of this type might eventually lead to similar theoretical insights as in the algebraic case In the following section we present the necessary background and notation in s dimensions In 3 we will present the well known class of 2-dimensional Fibonacci lattice rules and known results on their trigonometric degree In 4 we will investigate a particular extension of these rules to obtain an embedded pair and in 5 we compare their quality with what is theoretically the best possible result 2 A Short Course on Lattice Rules For a thorough introduction on lattice rules, we refer to [9] Results on the trigonometric degree of lattice rules up to that date were mainly published in the Russian literature and summarized in [1] Definition 1 A multiple integration lattice Λ in R s is a subset of R s which is discrete and closed under addition and subtraction and which contains Z s as a subset A lattice rule for approximating an integral over [0,1 s is a cubature formula where the N points are the points of a multiple integration lattice Λ that lie in [0,1 s and all points have the same weight 1/N Definition 2 The dual of the lattice Λ is Λ := {r R s : r x Z, x Λ} A lattice Λ can be specified by an s s matrix M known as a generator matrix, whose rows generate the lattice This means that all elements of Λ are of the form

5 Extensions of Fibonacci Lattice Rules 3 x = λm, where λ Z s The dual lattice Λ then has generator matrix B = (M 1 T Since Λ is an integration lattice, its dual Λ is an integer lattice and is generated by an integer-valued matrix B The dual of a multiple integration lattice plays an important role in the error representation and is the main tool to prove our results Assume that f can be expanded into an absolutely convergent multiple Fourier series f (x = a(re 2πir x with a(r = e 2πir x f (xdx, r Z s [0,1 s then the error is given by the next theorem Note that this assumption limits the functions to be 1-periodic in each dimension Theorem 1 [10] Let Λ be a multiple integration lattice Then the corresponding lattice rule Q has an error Q[ f ] I[ f ] = a(r r Λ \0 The trigonometric degree of a lattice rule can be determined from the dual lattice: d(q := min r Λ \0 r 1 There is a diamond shaped region (a crosspolytope to be precise with no points of the dual lattice except the origin inside, and some points on its boundary The 1-norm of points on the boundary is d + 1 Central symmetry (of the integration region and the points in the cubature formulas plays an important role in the algebraic case If this symmetry is present, then the cubature formula integrates the odd polynomials exactly automatically The lower bound for the number of points required is also fundamentally different for the even and the odd degrees The role that central symmetry plays in the algebraic case is played by shift symmetry in the trigonometric case [5] Definition 3 A cubature formula Q for an integral I on [0,1 s is shift symmetric if, whenever (x ( j j 1,,x( s is a point of the formula, then so is (x ( j j 2,,x( s + 1 2, with both points having the same weight The point (x ( j j,,x( s in the above theorem should actually be interpreted modulo 1, ie, wrapped around the edges of the unit cube However, since we are assuming periodic functions we loosen the notation and do not write the traditional fractional braces around the points to denote the modulo 1 From Definition 3 it follows that N is even for a shift symmetric cubature formula Furthermore, because a shift symmetric cubature formulas is automatically exact for all trigonometric monomials of odd degree, such a cubature formula has an odd trigonometric degree, see [5] Finally note that a lattice rule is shift symmetric if and only if ( 1 2,, 1 2 is a point of the lattice

6 4 Ronald Cools and Dirk Nuyens 3 Fibonacci Lattice Rules We will restrict our investigations in this paper to two dimensions, starting from a well known family of lattice rules Let F k be the kth Fibonacci number, defined by F 0 := 0, F 1 := 1 and F k := F k 1 + F k 2 for 1 < k N Consider the following lattice rules: L k [ f ] = 1 F k 1 ( j F k f, jf k 1 (1 j=0 F k F k and L k [ f ] = 1 F k 1 ( j F k f, jf k 2 j=0 F k F k Lattice rules of this form are called Fibonacci lattice rules The two rules given above are geometrically equivalent, meaning that one point set can be changed into the other one by symmetry operations of the unit cube [9] In this case a reflection on the second coordinate axis maps L k to L k since jf k 1 jf k 2 (mod F k Geometrically equivalent lattice rules have the same trigonometric degree, as is obvious from their dual lattices The kth Fibonacci number F k is even if and only if k is a multiple of 3 This can be observed by looking at the Fibonacci sequence modulo 2: (F k mod 2 k 1 = (1,1,0, 1,1,0, 1,1,0,, ie, the 2nd Pisano period is 3 Only in these cases are the Fibonacci lattice rules shift symmetric Indeed, the point of L k generated by j = F k /2 is 12, F k 1 2 and this then maps to ( 1 2, 1 2 The trigonometric degree of Fibonacci lattice rules is known explicitly Theorem 2 [1] If k = 2m+1 and m 2 then the Fibonacci lattice rule has trigonometric degree F m+2 1 If k = 2m then the Fibonacci lattice rule has trigonometric degree 2F m 1 Shift symmetric lattice rules have an odd trigonometric degree [5] The converse is not always true A rule of odd trigonometric degree is not necessarily shift symmetric The family of Fibonacci lattice rules has many examples of this The proof in [1] of the above theorem is based on the dual lattice We sketch it here because we will use the same technique in 4 and because it reveals a structure in the lattice Starting from the evident generator matrix, eg, for k = 2m + 1 M = ( 1 F 2m+1 F 2m F 2m it follows that B = (M 1 T F2m+1 0 = is a generator matrix for the dual lattice F 2m 1 When U is any unimodular matrix then UB is also a generator matrix (A unimodular matrix is a square ( integer matrix with determinant +1 or 1 Using the unimodular Fm F matrix U = m+1, it is shown that F m 1 F m

7 Extensions of Fibonacci Lattice Rules 5 ( Fm ( 1 B = UA with A = F m+1 F m+1 ( 1 m F m (2 Hence, A is also a generator matrix of this dual lattice Then, it is proven that no nonzero combination of the rows of this matrix leads to a point with 1-norm smaller than the claimed degree Observe that the generator matrix A (2 of the dual lattice is an orthogonal matrix (modulo F k = F 2m+1 = Fm 2 + Fm+1 2 In other words, both generating vectors are orthogonal and have the same length (in 2-norm So the Fibonacci lattice for odd k has a square unit cell, ie, this lattice corresponds to a rotated regular grid This fact was obtained in a different way, and explicitly recognized by Niederreiter and Sloan [8] 4 An Extension of Fibonacci Lattice Rules We are interested in lattice rules that extend Fibonacci lattice rules so that we obtain an embedded pair of lattice rules The aim is to obtain a pair that requires less function evaluations than two rules of the corresponding degrees We are partially successful in that respect Consider the lattice rule Q k [ f ] = 1 2F k 1 ( j 2F k f, jf k 1 (3 j=0 2F k 2F k This lattice is shift symmetric if F k 1 is odd Indeed for j = F k the point 12, F k 1 2 is generated Whenever F k 1 is odd, this maps to ( 1 2, 1 2 Next we will show that (3 can be seen as a Fibonacci lattice L k plus a shifted version of the same Fibonacci lattice The above rule can be split in two sums, separating even and odd values of j: Q k [ f ] = 1 F k 1 ( j 2F k f j=0 = 1 F k 1 2F k f j=0 F k, jf k 1 F k ( j F k, jf k 1 F k + 1 2F k + 1 F k 1 2F k f j=0 F k 1 ( 2 j + 1 f j=0 2F k, (2 j + 1F k 1 2F k ( j + 1, jf k 1 + F k 1 F k 2F k F k 2F k Thus the lattice rule (3 is given by the original rule (1 plus the original rule shifted by (1,F k 1 /(2F k (The weights are adjusted in the obvious way We will now show a formula for its trigonometric degree Theorem 3 The lattice rule given by (3 for k = 6m + α, with α = 0,1,,5, has trigonometric degree F 3m+α 1 + F 3m+1 1

8 6 Ronald Cools and Dirk Nuyens Proof The proof technique we use here was sketched at the end of 3 The cubature formula (3 with k = 6m + α is a lattice rule with generator matrix of the corresponding lattice: M = ( 1 2F 6m+α F 6m+α 1 2F 6m+α 0 1 We will instead investigate the minor modification of this matrix M = ( 1 2F 6m +α ( 1 m F 6m+α 1 2F 6m+α 0 ( 1 m If m is even, this is the same If m is odd, we investigate a geometrically equivalent lattice As mentioned before, geometrically equivalent lattice rules have the same trigonometric degree A generator matrix of the corresponding dual lattice is B = (M 1 T 2F6m+α 0 = F 6m+α 1 ( 1 m Observe that the matrix F3m+1 2F C = 3m F 3m /2 F 3m 1 is an integer unimodular matrix Indeed, we observed in 3 that F 3m is an even number and, making use of Cassini s identity, see (5 below, it follows that det(c = F 3m+1 F 3m 1 F 2 3m = ( 1 3m = ( 1 m We will first show that 2F3m+α 2F A = 3m F 3m+α 1 F 3m+1 (4 is another generator matrix for the dual lattice This follows from CA = B It requires some Fibonacci magic to show this The relevant relations for the sequel are [7] F k F n+1 + F k 1 F n = F n+k, F n+1 F n 1 F 2 n = ( 1 n (5 It follows that 2F3m+1 F CA = 3m+α + 2F 3m F 3m+α 1 2F 3m+1 F 3m 2F 3m F 3m+1 F 3m F 3m+α F 3m 1 F 3m+α 1 F3m 2 + F 3m 1F 3m+1 2F3m+1 F = 3m+α + 2F 3m F 3m+α 1 0 F 3m F 3m+α F 3m 1 F 3m+α 1 ( 1 3m = B

9 Extensions of Fibonacci Lattice Rules 7 The points on the dual lattice are generated as (i, ja = (i2f 3m+α jf 3m+α 1, i2f 3m + jf 3m+1 for all i, j Z The corresponding lattice rule has trigonometric degree F 3m+α 1 + F 3m+1 1 if all points of the dual lattice, except the point (0,0, lie outside or on the boundary of the diamond shaped region: i2f 3m+α jf 3m+α 1 + i2f 3m + jf 3m+1 F 3m+α 1 + F 3m+1 (6 (Some points fall exactly on the boundary, eg, when i = 0 and j = 1 We prove this inequality for different cases of (i, j For brevity we name parts of the inequality as follows β(α := i2f 3m+α jf 3m+α 1, δ := i2f 3m + jf 3m+1, γ(α := F 3m+α 1 + F 3m+1, so that we have to prove for all α = 0,1,,5: β(α + δ γ(α This is a tedious exercise in proving the bound for all different cases For reference the reader is referred to Fig 1 on different occasions during the proof, which depicts the possible integer values that the (i, j can take in generating the dual lattice points 1 If i = 0 then β(α + δ = j (F 3m+α 1 + F 3m+1 F 3m+α 1 + F 3m+1 since we must have j 1 2 If j = 0 then β(α + δ = i (2F 3m+α + 2F 3m F 3m+α 1 + F 3m+1 since we must have i 1 and 2F 3m F 3m + F 3m 1 = F 3m+1 3 If i j < 0 (ie, opposite signs then β(α = i 2F 3m+α + j F 3m+α 1 F 3m+α 1 + F 3m+1 since 2F 3m+α F 3m+α + F 3m+α 1 = F 3m+α+1 F 3m+1 for all α 0 We have now checked (and crossed out the 2nd and 4th quadrant, marked with i j 0 on Fig 1, to fulfill (6 4 For i j > 0 (ie, same signs we first look at β(α = i2f 3m+α jf 3m+α 1 = (2i jf 3m+α 1 + 2iF 3m+α 2 Again here we consider separate cases based on the signs of the two terms But note first that δ = i 2F 3m + j F 3m+1 2F 3m +F 3m+1, so we always trivially have the F 3m+1 term from γ(α and we can close a case quickly if β(α F 3m+α 1 Also note that we can get extra terms of F 3m and F 3m+1 by having larger bounds on respectively i and j a If 2i j = 0 then β(α = 2 i F 3m+α 2 2F 3m+α 2 F 3m+α 1 b If (2i ji > 0 (ie, same signs then β(α = 2i j F 3m+α 1 +2 i F 3m+α 2 F 3m+α 1

10 8 Ronald Cools and Dirk Nuyens ij 0 j 4 3 (2i ji i 1 2 (2i ji 0 3 ij 0 Fig 1 The different regions to consider for (i, j when taking linear combinations of the rows of the dual matrix A given by (4 At this time we have checked the largest parts of the 1st and 3rd quadrant, marked with (2i ji 0 on Fig 1, to fulfill (6 as well c If (2i ji < 0 (ie, opposite signs then we have more work First observe that from (2i ji < 0 we can conclude that j > 2, ie, j 3 (This is also visible on the figure We can thus refine our estimate for δ in this case to δ = i2f 3m + jf 3m+1 = i 2F 3m + j F 3m+1 2F 3m + 3F 3m+1 = 2F 3m+2 + F 3m+1 = F 3m+2 + F 3m+3 = F 3m+4 (The different expressions become useful in the following Now we consider the cases for the different values of α since we can easily obtain the required result by filling in the values, except for the case α = 5 where more work is needed

11 Extensions of Fibonacci Lattice Rules 9 i If α = 0,1,2,3,4 then we get the following results: α γ(α = F 3m+α 1 + F 3m+1 0 F 3m 1 + F 3m+1 1 F 3m + F 3m+1 = F 3m+2 2 F 3m+1 + F 3m+1 = 2F 3m+1 3 F 3m+2 + F 3m+1 = F 3m+3 4 F 3m+3 + F 3m+1 = F 3m+2 + 2F 3m+1 We note that for all of these values of α we have that δ γ(α (using the refined estimate for δ for all conditions we have set ii If α = 5 then β(5 = i2f 3m+5 jf 3m+4, δ = F 3m+4, γ(5 = F 3m+4 + F 3m+1, where we thankfully use the refined value of δ to only have to show β(5 F 3m+1 We can rewrite β(5 as follows β(5 = i2f 3m+5 jf 3m+4 = (10i 3 jf 3m+1 + 2(3i jf 3m Set a = 10i 3 j and b = 3i j and consider the different sign settings A If a = 0 then we need integer solutions of i = 10 3 j This means j 10 (and i 3 and we can make a new refinement for δ, δ 6F 3m + 10F 3m+1, which solves this case since we have (2F 3m + 3F 3m+1 + F 3m+1 = F 3m+4 + F 3m+1 B If b = 0 then we obtain j = 3i and as such a = 10i 9i = i from which it follows that β(5 = F 3m+1 Ie, for i = 1 and j = 3 we get a point on the boundary C If ab > 0 then the result is trivial as clearly then β(5 F 3m+1 +2F 3m We now have checked the darkest marked part on Fig 1 to fulfill (6 and are left with the narrow white wedge in between j = 10 3 i and j = 3i D If ab < 0 then we either have { { 10i 3 j < 0 10i 3 j > 0 or 6i 2 j > 0 6i 2 j < i < j < 3i 3i < j < 3 i From the inequality on the left side follows that both i and j should be negative, while the inequality on the right side makes it that both i and j are positive Combined we can write: 3 i < j < 10 3 i

12 10 Ronald Cools and Dirk Nuyens This means j 4 (and is easily checked on the figure Refining the estimate of δ we get and thus δ γ(5 This completes our proof δ = i 2F 3m + j F 3m+1 2F 3m + 4F 3m+1 = (2F 3m + 3F 3m+1 + F 3m+1 = F 3m+4 + F 3m+1, The result of the theorem is put together in Table 1 It is clear from the table that the cases for α = 1,2,3 are not interesting The extended rules have the same degree as the original rules, and the number of additional points is the same as for the original rule In other words, using the embedded rules costs the same as using two rules In the next section we will show that the remaining cases are however interesting Table 1 Trigonometric degrees of Q k for the different cases of Theorem 3 compared to the degree of the basic rule L k k mod 6 k mod 2 k mod 3 d(l k d(q k comparison F 3m 1 2F 3m + F 3m 3 1 higher F 3m+2 1 F 3m+2 1 equal F 3m+1 1 2F 3m+1 1 equal F 3m+3 1 F 3m+3 1 equal F 3m+2 1 2F 3m+2 + F 3m 1 1 higher F 3m+4 1 F 3m+4 + F 3m+1 1 higher Note that the line marked with a star ( includes k = 6, ie, m = 1 and α = 0 In this exceptional case d(l k = d(q k since then F 3m 3 = F 0 = 0 For all other values of k 0 (mod 6 the rule Q k does have a higher degree than the basic rule L k For the double starred line ( there is no such problem if one allows negative indices for the Fibonacci numbers (see, eg, [7] For k = 4, ie, m = 0 and α = 4, we then have d(q 4 = F 3 + F 1 1 = 2, while d(l 4 = 2F 2 1 = 1 Initially, we recognised the five different cases, and we attempted proofs for each α using a generator matrix that suited each case best There are indeed other useful generator matrices of the dual lattice We can, eg, rewrite the matrix A used in the previous proof: 2F3m+α 2F A = 3m = F 3m+α 1 F 3m+1 This can be transformed as follows 1 1 A = 0 1 ( F3m+2+α F 3m 1+α F 3m+2 F 3m 1 F 3m+α 1 F 3m+1 ( F3m+2+α F 3m 2 F 3m+α 1 F 3m+1

13 Extensions of Fibonacci Lattice Rules 11 For specific values of α this generator matrix is more convenient Of special interest is the case α = 5 It causes extra difficulties in the general proof given above If treated separately, that part of the proof becomes easier The difficult part is to show that the generator matrix B can be transformed to F3m+1 F A = 3m+α 1 F 3m+α 1 F 3m+1 by a unimodular matrix if α = 5 With hindsight, this matrix was guessed after the general proof was established Once this is established, proving the degree of the lattice rule is straightforward This case is special because now there are four points (instead of two of the dual lattice, one in each quadrant, lying on the boundary This lattice has a square unit cell This special structure is immediately evident from the matrix given above, but not from the matrix used in the general proof 5 Final Remarks Not all pairs of embedded cubature rules are interesting from a practical point of view Let N opt d denote the number of points used by an optimal rule of degree d, ie, one with the lowest number of points If we have two embedded rules of respectively degree d 1 and d 2 with N the total number of points, and d 1 < d 2, then a measure for its quality is N γ := N opt d 1 + N opt d 2 Obviously, we prefer rules with γ < 1 Lower bounds on the number of points needed to achieve a given trigonometric degree are known (see, eg, [1, 5, 4] In two dimensions the lower bound for a degree d rule is given by N N opt d := { 2m 2 + 2m + 1, for d = 2m, 2(m + 1 2, for d = 2m + 1 Furthermore, lattice rules are known that attain this lower bound (see, eg, [1, 5] In Table 2 it can be seen that for the cases k = 6m, 6m + 4 and 6m + 5 investigated above, we have that γ < 1 The data in Table 2 was computationally verified completely up to k = 54 Acknowledgements The authors would like to thank Stephen Joe for commenting on a draft version of the manuscript The second author acknowledges support of the Australian Research Council and the Research Foundation Flanders This paper presents research results of the Belgian Network DYSCO (Dynamical Systems, Control, and Optimization, funded by the Interuniversity Attraction Poles Programme, initiated by the Belgian State, Science Policy Office The scientific responsibility rests with its author(s

14 12 Ronald Cools and Dirk Nuyens Table 2 Comparison for the trigonometric degrees of the embedded rule Q k based on the Fibonacci rule L k with the lower bounds for the same degrees Here α = k mod 6, d 1 = d(l k, N = 2F k and d 2 = d(q k For k = 6 the value of γ should be ignored since there d 1 = d 2 k α F k d 1 N opt d 1 N d 2 N opt d 2 N opt d 1 + N opt d γ References 1 Beckers, M, Cools, R: A relation between cubature formulae of trigonometric degree and lattice rules In: H Brass, G Hämmerlin (eds Numerical Integration IV, pp Birkhäuser Verlag, Basel ( Cools, R, Haegemans, A: Optimal addition of knots to cubature formulae for planar regions Numer Math 49, ( Cools, R, Haegemans, A: A lower bound for the number of function evaluations in an error estimate for numerical integration Constr Approx 6, ( Cools, R, Nuyens, D: A Belgian view on lattice rules In: A Keller, et al (eds Monte Carlo and Quasi-Monte Carlo Methods 2006, pp 3 21 Springer ( Cools, R, Sloan, I: Minimal cubature formulae of trigonometric degree Math Comp 65(216, ( Davis, P, Rabinowitz, P: Methods of Numerical Integration Academic Press, London ( Graham, RL, Knuth, DE, Patashnik, O: Concrete Mathematics, 2nd edn Addison-Wesley ( Niederreiter, H, Sloan, I: Integration of nonperiodic functions of two variables by Fibonacci lattice rules J Comput Appl Math 51, ( Sloan, I, Joe, S: Lattice Methods for Multiple Integration Oxford University Press ( Sloan, I, Kachoyan, P: Lattice mathods for multiple integration: theory, error analysis and examples SIAM J Numer Anal 24, (1987